a cup of coffee at 95 degrees Celsius is put into a 25 degree Celsius room when t=0. The coffee's temperature,f(t), is changing at a rate given by f'(t)=-5(0.6)^t degrees Celsius per minute, where t is in minutes. Estimate the coffee's temperature when t=10.

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Here is one way to go about it.

Newton's Cooling Law says that, in this case:

\frac{dT}{dt}=k(T-25)

\frac{dT}{T-25}=kdt

Integrate and get:

ln(T-25)=kt+C

But, at t=0, the coffee is T=95.

Therefore, subbing in T=95 and t=0, we find that C=ln(70).

But, we are also told that

k=-5(\frac{3}{5})^{t}

So, we have:

ln(\frac{T-25}{70})=-5t(\frac{3}{5})^{t}

Sub in t=10 and solve for T to find the coffee temp. at t=10.